The area of the region bounded by the curve y=lvert x^2-4rvert and the line y=5 is

The line meets the curve where $|x^2-4|=5$. Since $x^2-4=-5$ has no real solution, the only intersections are $x^2=9$, i.e. $x=\pm3$. The region is symmetric about the $y$-axis, so area $=2\int_0^3\bigl(5-|x^2-4|\bigr)dx$. On $[0,2]$, $|x^2-4|=4-x^2$, so the integrand is $1+x^2$ and $\int_0^2(1+x^2)dx=2+\frac83=\frac{14}3$. On $[2,3]$, $|x^2-4|=x^2-4$, so the integrand is $9-x^2$ and $\int_2^3(9-x^2)dx=9-\frac{19}3=\frac83$. Hence the area is $2\left(\frac{14}3+\frac83\right)=\frac{44}3$. The value $36$ results from ignoring the modulus and integrating $9-x^2$ over $[-3,3]$.